# Trigonometric Integrals

In this topic, we will study how to integrate certain combinations involving products and powers of trigonometric functions.

We consider $$8$$ cases.

### 1. Integrals of the form $${\large\int\normalsize} {\cos ax\cos bxdx} ,$$ $${\large\int\normalsize} {\sin ax\cos bxdx} ,$$ $${\large\int\normalsize} {\sin ax\sin bxdx}$$

To evaluate integrals of products of sine and cosine with different arguments, we apply the identities

• $${\cos ax\cos bx }$$ $$= {{\large\frac{1}{2}\normalsize} \left[ {\cos \left( {ax + bx} \right) }\right.}$$ $$+\, {\left.{ \cos \left( {ax – bx} \right)} \right]}$$
• $${\sin ax\cos bx }$$ $$= {{\large\frac{1}{2}\normalsize} \left[ {\sin \left( {ax + bx} \right) }\right.}$$ $$+\, {\left.{ \sin \left( {ax – bx} \right)} \right]}$$
• $${\sin ax\sin bx }$$ $$= {-{\large\frac{1}{2}\normalsize} \left[ {\cos \left( {ax + bx} \right) }\right.}$$ $$-\, {\left.{ \cos \left( {ax – bx} \right)} \right]}$$

### 2. Integrals of the form $${\large\int\normalsize} {{\sin^m}x\,{\cos^n}xdx}$$

We assume here that the powers $$m$$ and $$n$$ are non-negative integers.

To find an integral of this form, use the following substitutions:

1. If $$m$$ (the power of sine) is odd, we use the $$u-$$substitution ${u = \cos x,\;\;}\kern0pt{du = – \sin xdx}$ and the identity $${\sin ^2}x + {\cos ^2}x = 1$$ to express the remaining even power of sine in $$u-$$terms.
2. If $$n$$ (the power of cosine) is odd, we use the $$u-$$substitution ${u = \sin x,\;\;}\kern0pt{du = \cos xdx}$ and the identity $${\sin ^2}x + {\cos ^2}x = 1$$ to express the remaining even power of cosine in $$u-$$terms.
3. If both powers $$m$$ and $$n$$ are even, we reduce the powers using the the double angle formulas ${{\sin ^2}x = \frac{{1 – \cos 2x}}{2},\;\;}\kern0pt{{\cos ^2}x = \frac{{1 + \cos 2x}}{2}.}$

The integrals of type $$\int {{{\sin }^n}xdx}$$ and $$\int {{{\cos }^n}xdx}$$ can be evaluated by reduction formulas

${\int {{{\sin }^n}xdx} }={ – \frac{{{{\sin }^{n – 1}}x\cos x}}{n} }+{ \frac{{n – 1}}{n}\int {{{\sin }^{n – 2}}xdx} ,}$

${\int {{{\cos }^n}xdx} }={ \frac{{{{\cos }^{n – 1}}x\sin x}}{n} }+{ \frac{{n – 1}}{n}\int {{{\cos }^{n – 2}}xdx} .}$

### 3. Integrals of the form $${\large\int\normalsize} {{\tan^n}xdx}$$

The power of the integrand can be reduced using the trigonometric identity $$1 + {\tan ^2}x = {\sec ^2}x$$ and the reduction formula

${\int {{\tan^n}xdx} } = {\int {{{\tan }^{n – 2}}x\,{{\tan }^2}xdx} } = {\int {{{\tan }^{n – 2}}x\left( {{{\sec }^2}x – 1} \right)dx} } = {\frac{{{{\tan }^{n – 1}}x}}{{n – 1}} }-{ \int {{{\tan }^{n – 2}}xdx} .}$

### 4. Integrals of the form $${\large\int\normalsize} {{{\cot }^n}xdx}$$

The power of the integrand can be reduced using the trigonometric identity $$1 + {\cot ^n}x = {\csc ^n}x$$ and the reduction formula

${\int {{{\cot }^n}xdx} } = {\int {{{\cot }^{n – 2}}x\,{{\cot }^2}xdx} } = {\int {{{\cot }^{n – 2}}x\left( {{{\csc }^2}x – 1} \right)dx} } = { – \frac{{{{\cot }^{n – 1}}x}}{{n – 1}} }-{ \int {{{\cot }^{n – 2}}xdx} .}$

### 5. Integrals of the form $${\large\int\normalsize} {{\sec^n}xdx}$$

This type of integrals can be simplified with help of the reduction formula:

${\int {{\sec^n}xdx} } = {\frac{{{{\sec }^{n – 2}}x\tan x}}{{n – 1}} }+{ \frac{{n – 2}}{{n – 1}}\int {{\sec^{n – 2}}xdx} .}$

### 6. Integrals of the form $${\large\int\normalsize} {{\csc^n}xdx}$$

Similarly to the previous examples, this type of integrals can be simplified by the formula

${\int {{\csc^n}xdx} } = { – \frac{{{\csc^{n – 2}}x \cot x}}{{n – 1}} }+{ \frac{{n – 2}}{{n – 1}}\int {{\csc^{n – 2}}xdx} .}$

### 7. Integrals of the form $${\large\int\normalsize} {{\tan^m}x\,{\sec^n}xdx}$$

1. If the power of the secant $$n$$ is even, then using the identity $$1 + {\tan ^2}x$$ $$= {\sec ^2}x$$ the secant function is expressed as the tangent function. The factor $${\sec ^2}x$$ is separated and used for transformation of the differential. As a result, the entire integral (including differential) is expressed in terms of the function $$\tan x.$$
2. If both the powers $$n$$ and $$m$$ are odd, then the factor $$\sec x \tan x,$$ which is necessary to transform the differential, is separated. Then the entire integral is expressed in terms of $$\sec x.$$
3. If the power of the secant $$n$$ is odd, and the power of the tangent $$m$$ is even, then the tangent is expressed in terms of the secant using the identity $$1 + {\tan ^2}x$$ $$= {\sec ^2}x.$$ After this substitution, you can calculate the integrals of the secant.

### 8. Integrals of the form $${\large\int\normalsize} {{\cot^m}x\,{\csc^n}xdx}$$

1. If the power of the cosecant $$n$$ is even, then using the identity $$1 + {\cot^2}x = {\csc ^2}x$$ the cosecant function is expressed as the cotangent function. The factor $${\csc^2}x$$ is separated and used for transformation of the differential. As a result, the integrand and differential are expressed in terms of $$\cot x.$$
2. If both the powers $$n$$ and $$m$$ are odd, then the factor $$\cot x \csc x,$$ which is necessary to transform the differential, is separated. Then the integral is expressed in terms of $$\csc x.$$
3. If the power of the cosecant $$n$$ is odd, and the power of the cotangent $$m$$ is even, then the cotangent is expressed in terms of the cosecant using the identity $$1 + {\cot^2}x = {\csc ^2}x.$$ After this substitution, you can find the integrals of the cosecant.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Calculate the integral $${\large\int\normalsize} {{\sin^3}xdx}.$$

### Example 2

Evaluate the integral $${\large\int\normalsize} {{\cos^5}xdx}.$$

### Example 3

Find the integral $${\large\int\normalsize} {{\sin^6}xdx}.$$

### Example 4

Find the integral $$\int {{{\sin }^2}x\,{{{\cos }^3}x}dx}.$$

### Example 5

Calculate the integral $${\large\int\normalsize} {{{\sin }^2}x\,{{\cos }^4}xdx}.$$

### Example 6

Evaluate the integral $$\int {{{\sin }^3}x\,{{\cos }^4}xdx}.$$

### Example 7

Evaluate the integral $$\int {{{\sin }^3}x\,{{\cos }^5}xdx}.$$

### Example 8

Evaluate the integral $$\int {{{\sin }^3}x\,{{\cos }^3}xdx}.$$

### Example 9

Find the integral $${\large\int\normalsize} {{{\sin }^3}x\sqrt {\cos x} dx}.$$

### Example 10

Evaluate the integral $$\int {\sin 2x\cos 5xdx}.$$

### Example 11

Evaluate the integral $${\large\int\normalsize} {\sin {\large\frac{x}{4}\normalsize} \cos {\large\frac{x}{3}\normalsize} dx}.$$

### Example 12

Evaluate the integral $$\int {\cos \large{\frac{x}{2}}\normalsize\cos \large{\frac{x}{3}}\normalsize dx}.$$

### Example 13

Evaluate the integral $$\int {\sin \large{\frac{x}{3}}\normalsize\sin \large{\frac{x}{4}}\normalsize dx}.$$

### Example 14

Evaluate the integral $${\large\int\normalsize} {{{\tan }^4}xdx}.$$

### Example 15

Find the integral $$\int {{{\cot }^4}xdx}.$$

### Example 16

Find the integral $$\int {{{\tan }^5}xdx}.$$

### Example 17

Calculate the integral $${\large\int\normalsize} {{{\cot }^5}xdx}.$$

### Example 18

Calculate the integral $${\large\int\normalsize} {{{\sec }^3}xdx}.$$

### Example 19

Find the integral $$\int {{{\csc }^3}xdx}.$$

### Example 20

Evaluate the integral $${\large\int\normalsize} {{{\csc }^4}xdx}.$$

### Example 21

Find the integral $$\int {{{\sec }^4}xdx}.$$

### Example 22

Compute $${\large\int\normalsize} {{\tan^3}x\,{{\sec }^2}xdx}.$$

### Example 23

Compute $${\large\int\normalsize} {{\tan^2}x\sec xdx}.$$

### Example 1.

Calculate the integral $${\large\int\normalsize} {{\sin^3}xdx}.$$

Solution.

Let $$u = \cos x,$$ $$du = -\sin xdx.$$ Then

${\int {{\sin^3}xdx} }={ \int {{\sin^2}x\sin xdx} } = {\int {\left( {1 – {\cos^2}x} \right)\sin xdx} } = { – \int {\left( {1 – {u^2}} \right)du} } = {\int {\left( {{u^2} – 1} \right)du} } = {\frac{{{u^3}}}{3} – u + C } = {{\frac{{{{\cos }^3}x}}{3} – \cos x }+{ C.}}$

### Example 2.

Evaluate the integral $${\large\int\normalsize} {{\cos^5}xdx}.$$

Solution.

Making the substitution $$u = \sin x,$$ $$du = \cos xdx$$ and using the identity $${\cos ^2}x = 1 – {\sin ^2}x,$$ we obtain

${\int {{\cos^5}xdx} } = {\int {{{\left( {{\cos^2}x} \right)}^2}\cos xdx} } = {\int {{{\left( {1 – {{\sin }^2}x} \right)}^2}\cos x dx} } = {\int {{{\left( {1 – {u^2}} \right)}^2}du} } = {\int {\left( {1 – 2{u^2} + {u^4}} \right)du} } = {{u – \frac{{2{u^3}}}{3} + \frac{{{u^5}}}{5} }+{ C }} = {{\sin x – \frac{{2{{\sin }^3}x}}{3} }+{ \frac{{{{\sin }^5}x}}{5} }+{ C.}}$

### Example 3.

Find the integral $${\large\int\normalsize} {{\sin^6}xdx}.$$

Solution.

Using identities $${\sin ^2}x = {\large\frac{{1 – \cos 2x}}{2}\normalsize}$$ and $${\cos ^2}x = {\large\frac{{1 + \cos 2x}}{2}\normalsize},$$ we can write:

${I = \int {{\sin^6}xdx} } = {\int {{{\left( {{\sin^2}x} \right)}^3}dx} } = {\frac{1}{8}\int {{{\left( {1 – \cos 2x} \right)}^3}dx} } = {{\frac{1}{8}\int {\left( {1 – 3\cos 2x }\right.}}}+{{{\left.{ 3{{\cos }^2}2x – {{\cos }^3}2x} \right)dx} }} = {\frac{x}{8} – \frac{3}{8} \cdot \frac{{\sin 2x}}{2} } + {\frac{3}{8}\int {{\cos^2}2xdx} } – {\frac{3}{8}\int {{\cos^3}2xdx}.}$

Calculate the integrals in the latter expression.

${\int {{\cos^2}2xdx} } = {\int {\frac{{1 + \cos 4x}}{2}dx} } = {\frac{1}{2}\int {\left( {1 + \cos 4x} \right)dx} } = {\frac{1}{2}\left( {x + \frac{{\sin 4x}}{4}} \right) } = {\frac{x}{2} + \frac{{\sin 4x}}{8}.}$

To find the integral $${\large\int\normalsize} {{\cos^3}2xdx},$$ we make the substitution $$u = \sin 2x,$$ $$du =$$ $$2\cos 2xdx.$$ Then

${\int {{\cos^3}2xdx} } = {\frac{1}{2}\int {2{{\cos }^2}2x\cos 2xdx} } = {\frac{1}{2}\int {2\left( {1 – {{\sin }^2}2x} \right)\cos 2xdx} } = {\frac{1}{2}\int {\left( {1 – {u^2}} \right)du} } = {\frac{u}{2} – \frac{{{u^3}}}{6} } = {\frac{{\sin 2x}}{2} – \frac{{{{\sin }^3}2x}}{6}.}$

Hence, the initial integral is

${I = \frac{x}{8} – \frac{{3\sin 2x}}{{16}} } + {\frac{3}{8}\left( {\frac{x}{2} + \frac{{\sin 4x}}{8}} \right) } – {\frac{1}{8}\left( {\frac{{\sin 2x}}{2} – \frac{{{{\sin }^3}2x}}{6}} \right) }+{ C } = {\frac{{5x}}{{16}} – \frac{{\sin 2x}}{4} } + {\frac{{3\sin 4x}}{{64}} } + {\frac{{{{\sin }^3}2x}}{{48}} }+{ C.}$

### Example 4.

Find the integral $$\int {{{\sin }^2}x\,{{{\cos }^3}x}dx}.$$

Solution.

The power of cosine is odd, so we make the substitution

${u = \sin x,\;\;}\kern0pt{du = \cos xdx.}$

We rewrite the integral in terms of $$\sin x$$ to obtain:

${\int {{{\sin }^2}x\,{{\cos }^3}xdx} }={ \int {{{\sin }^2}x\,{{{\cos }^2}x}\cos xdx} }={ \int {{{\sin }^2}x{\left( {1 – {{\sin }^2}x} \right)}\cos xdx} }={ \int {{u^2}\left( {1 – {u^2}} \right)du} }={ \int {\left( {{u^2} – {u^4}} \right)du} }={ \frac{{{u^3}}}{3} – \frac{{{u^5}}}{5} + C }={ \frac{{{{\sin }^3}x}}{3} – \frac{{{{\sin }^5}x}}{5} + C.}$

### Example 5.

Calculate the integral $${\large\int\normalsize} {{{\sin }^2}x\,{{\cos }^4}xdx}.$$

Solution.

We can write:

${I = \int {{{\sin }^2}x\,{{\cos }^4}xdx} } = {\int {{{\left( {\sin x\cos x} \right)}^2}{{\cos }^2}xdx} .}$

We convert the integrand using the identities

${\sin x\cos x = \frac{{\sin 2x}}{2},\;\;\;}\kern-0.3pt {{\cos ^2}x = \frac{{1 + \cos 2x}}{2},\;\;\;}\kern-0.3pt {{\sin ^2}x = \frac{{1 – \cos 2x}}{2}.}$

This yields

${I = \int {{{\left( {\frac{{\sin 2x}}{2}} \right)}^2}\frac{{1 + \cos 2x}}{2}dx} } = {\frac{1}{8}\int {{{\sin }^2}2x\left( {1 + \cos 2x} \right)dx} } = {{\frac{1}{8}\int {{{\sin }^2}2xdx} }+{ \frac{1}{8}\int {{{\sin }^2}2x\cos 2xdx} }} = {{\frac{1}{8}\int {\frac{{1 – \cos 4x}}{2}dx} }}+{{ \frac{1}{{16}}\int {2{{\sin }^2}2x\cos 2xdx} }} = {{\frac{1}{{16}}\int {\left( {1 – \cos 4x} \right)dx} }}+{{ \frac{1}{{16}}\int {{{\sin }^2}2x\,d\left( {\sin 2x} \right)} }} = {{\frac{1}{{16}}\left( {x – \frac{{\sin 4x}}{4}} \right) }}+{{ \frac{1}{{16}} \cdot \frac{{{{\sin }^3}2x}}{3} }+{ C }} = {{\frac{x}{{16}} – \frac{{\sin 4x}}{{64}} }+{ \frac{{{{\sin }^3}2x}}{{48}} }+{ C.}}$

### Example 6.

Evaluate the integral $$\int {{{\sin }^3}x\,{{\cos }^4}xdx}.$$

Solution.

As the power of sine is odd, we use the substitution

${u = \cos x,\;\;}\kern0pt{du = – \sin xdx.}$

The integral is written as

${I = \int {{{\sin }^3}x\,{{\cos }^4}xdx} }={ \int {{{\sin }^2}x\,{{\cos }^4}x\sin xdx} .}$

By the Pythagorean identity,

${\sin ^2}x = 1 – {\cos ^2}x.$

Hence

${I = \int {{{\sin }^2}x\,{{\cos }^4}x\sin xdx} }={ \int {\left( {1 – {{\cos }^2}x} \right){{\cos }^4}x\sin xdx} }={ – \int {\left( {1 – {u^2}} \right){u^4}du} }={ \int {\left( {{u^6} – {u^4}} \right)du} }={ \frac{{{u^7}}}{7} – \frac{{{u^5}}}{5} + C }={ \frac{{{{\cos }^7}x}}{7} – \frac{{{{\cos }^5}x}}{5} + C.}$

### Example 7.

Evaluate the integral $$\int {{{\sin }^3}x\,{{\cos }^5}xdx}.$$

Solution.

We see that both powers are odd, so we can substitute either $$u = \sin x$$ or $$u = \cos x.$$ Choosing the least exponent, we have

${u = \cos x,\;\;}\kern0pt{du = – \sin xdx.}$

The integral takes the form

${I = \int {{{\sin }^3}x\,{{\cos }^5}xdx} }={ \int {{{\sin }^2}x\,{{\cos }^5}x\sin xdx} .}$

Using the Pythagorean identity,

${\sin ^2}x = 1 – {\cos ^2}x,$

we can write

${I = \int {{{\sin }^2}x\,{{\cos }^5}x\sin xdx} }={ \int {\left( {1 – {{\cos }^2}x} \right){{\cos }^5}x\sin xdx} }={ – \int {\left( {1 – {u^2}} \right){u^5}du} }={ \int {\left( {{u^7} – {u^5}} \right)du} }={ \frac{{{u^8}}}{8} – \frac{{{u^6}}}{6} + C }={ \frac{{{{\cos }^8}x}}{8} – \frac{{{{\cos }^6}x}}{6} + C.}$

### Example 8.

Evaluate the integral $$\int {{{\sin }^3}x\,{{\cos }^3}xdx}.$$

Solution.

The powers of both sine and cosine are odd. Hence we can use the substitution $$u = \sin x$$ or $$u = \cos x.$$ Let’s apply the substitution $$u = \sin x.$$ Then $$du = \cos x dx,$$ and the integral becomes

${I = \int {{{\sin }^3}x\,{{\cos }^3}xdx} }={ \int {{{\sin }^3}x\,{{\cos }^2}x\cos xdx} .}$

By the Pythagorean identity,

${\sin ^2}x = 1 – {\cos ^2}x,$

so we obtain

${I = \int {{{\sin }^3}x\,{{\cos }^2}x\cos xdx} }={ \int {{{\sin }^3}x\left( {1 – {{\sin }^2}x} \right)\cos xdx} }={ \int {{u^3}\left( {1 – {u^2}} \right)du} }={ \int {\left( {{u^3} – {u^5}} \right)du} }={ \frac{{{u^4}}}{4} – \frac{{{u^6}}}{6} + C }={ \frac{{{{\sin }^4}x}}{4} – \frac{{{{\sin }^6}x}}{6} + C.}$

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Problems 9-23